Questions 1
(a)Topic -5.1 Enthalpy change, ΔH
(b)Topic -5.1 Enthalpy change, ΔH
(c)Topic -10.1 Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
(a) The most common zinc mineral contains zinc(II) sulfide, ZnS.
(i) Complete the electrons in boxes diagram in Fig. 1.1 to show the electronic configuration of a zinc(II) ion.
(ii) Complete Fig. 1.2 to show the Born–Haber diagram for the ionic solid ZnS. Include state symbols of relevant species 
(iii) Describe the trend in the first electron affinity of the Group 16 elements S to Te. Explain your answer.
(iv) Explain why the lattice energy, \(\Delta H_{latt}\), of ZnO is more exothermic than that of ZnS.
(b) Zinc metal can be obtained in a two-step process as shown.
step 1 2ZnS(s) + 3O₂(g) \(\to\) 2ZnO(s) + 2SO₂(g)
step 2 ZnO(s) + C(s) \(\to\) Zn(l) + CO(g)
The reactions are carried out at 800°C.
(i) Predict the sign of the entropy change, \(\Delta S^0\) , of the reaction in step 1. Explain your answer
(ii) Use the data in Table 1.1 to calculate \(\Delta S^0\) of the reaction shown in step 2.
(iii) An equation for the direct reduction of ZnS by carbon is shown.
\(2ZnS(s) + C(s) \to 2Zn(l) + CS_2(g)\) \(\Delta H^0= +733kJmol^{–1}\)
\(\Delta S^0= +218JK^{–1}mol^{–1}\)
This reaction is not feasible at 800°C. Calculate \(\Delta G^0\) for this reaction at 800°C.
(c) Zn(NO₃)₂ undergoes thermal decomposition when heated. The reaction is similar to the thermal decomposition of Group 2 nitrates.
(i) Construct an equation for the thermal decomposition of Zn(NO₃)₂.
(ii) The radii of some Group 2 cations and Zn²⁺ are shown in Table 1.2.
State and explain the trend in thermal stability of the Group 2 nitrates down the group.
(iii) Use Table 1.2 to suggest which Group 2 nitrates are less thermally stable than zinc nitrate.
▶️Answer/Explanation
Ans:
(iii) • EA becomes less negative/ less exothermic (down group / S to Te)
• atomic radii increases OR outer shell gets farther from nucleus OR electron added at higher energy level
OR more shielding (of outer shells)
• less nuclear attraction
OR less attraction for incoming/added electron
any two [1] all three [2]
(iv) M1: \(O^{2–}\) (has same charge but) smaller (radius than \(S^{2–}\)) ORA
OR oxygen has a smaller ion (than \(S^{2–}\))
M2: stronger ionic bond OR greater attraction between Zn²⁺ and \(O^{2–}\) ORA
(b)(i) ΔS negative AND more moles / molecules of gaseous reactants ORA
OR ΔS is negative AND moles / molecules of gas are reduced (in the reaction)
(ii) ΔS = 50.8 + 197.7 – 43.7 – 5.7 = (+)199.1 (J K⁻¹ mol⁻¹)
(iii) ΔG = ΔH – TΔS ALLOW G = ΔH – TΔS
= +733 – (800 + 273) \(\times\) 0.218 = (+)499.086 (kJ mol⁻¹) min 3sf
(c)(i) Zn(NO₃)₂ → ZnO + 2NO₂ + ½O₂
OR 2Zn(NO₃)₂ → 2ZnO + 4NO₂ + O₂
(ii) increases (in thermal stability down the group)
AND (cat)ion(ic) radius / ion size increases (down the group)
less polarisation / less distortion of anion/ of nitrate ion/\(NO_3^–/ NO_3^{2–}\)
OR less weakening of N—O bond
(iii) Mg(NO₃)₂ only ALLOW Mg²⁺ / magnesium
Questions 2
(a)Topic -6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
(b)Topic -6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
(c)Topic -8.1 Rate of reaction
(d)Topic -8.1 Rate of reaction
Hypophosphorous acid is an inorganic acid. The conjugate base of hypophosphorous acid is H₂PO₂⁻.
(a) Give the formula of hypophosphorous acid.
(b) H₂PO₂⁻ is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis.
equation 1 \(HPO_3^{2–} + 2H_2O + 2e^– \rightleftharpoons H_2PO_2^– + 3OH^–\) \(E^o = –1.57V\)
(i) In an experiment, an alkaline HPO₃²⁻/H₂PO₂⁻ half-cell is constructed with [H₂PO₂⁻] = 0.050 mol dm⁻³. All other ions are at their standard concentration. Predict how the value of E of this half-cell differs from its E o value. Explain your answer.
(ii) The Cr³⁺/Cr half-cell has a standard electrode potential of –0.74V. An electrochemical cell consists of an alkaline HPO₃²⁻/H₂PO₂⁻ half-cell and a Cr³⁺/Cr half-cell. Calculate the standard cell potential, Ecell.
(iii) Complete the diagram in Fig. 2.1 to show how the standard electrode potential of the Cr³⁺/Cr half-cell can be measured relative to that of the standard hydrogen electrode. Identify the chemicals, conditions and relevant pieces of apparatus.
(iv) Label Fig. 2.1 to show:
• which is the positive electrode
• the direction of electron flow in the external circuit.
(v) \(H_2PO_2^–\) reduces \(Ni^{2+}\) to Ni in alkaline conditions. Use equation 1 to construct the ionic equation for this reaction.
equation 1 \(HPO_3^{2–} + 2H_2O + 2e^– \rightleftharpoons H_2PO_2^– + 3OH^–\)
(c) \(H_2PO_2^–\)(aq) reacts with \(OH^–\)(aq).
\(H_2PO_2^–(aq) + OH^–(aq) \to HPO_3^{2–}(g) + H_2(g)\)
Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.
(i) The volume of H₂ was measured under room conditions. Use the molar volume of gas, \(V_m\), and the data from experiment 1 to calculate the rate of reaction in \(moldm^{–3} s^{–1}\).
(ii) The rate equation was found to be:
\(rate = k [H_2PO_2^-(aq)] [OH^–(aq)]^2\)
Show that the data in Table 2.1 is consistent with the rate equation.
(iii) State the units of the rate constant, k, for the reaction.
(iv) The experiment is repeated using a large excess of OH⁻(aq). Under these conditions, the rate equation is:
rate = k₁ [H₂PO₂⁻(aq)]
k₁ = 8.25 × 10⁻⁵ s⁻¹
Calculate the value of the half-life, \(t_\frac{1}{2}\), of the reaction.
(v) Describe how an increase in temperature affects the value of the rate constant, k₁.
(d) A student suggests that the reaction between H₂PO₂⁻(aq) and OH–(aq) might happen more quickly in the presence of a heterogeneous catalyst. Describe the mode of action of a heterogeneous catalyst.
▶️Answer/Explanation
Ans:
(a) \(H_3PO_2\)
(b)(i) electrode potential E would become more positive / less negative (than \(E^0\))
lower [H₂PO₂⁻] AND shifts equilibrium to the right-hand side
(ii) +1.57 – 0.74 = (+)0.83 (V)
(iv) Pt electrode positive AND flow of electrons anticlockwise (to the SHE)
(v) H₂PO₂⁻ + 3OH⁻ + Ni²⁺ → HPO₃²⁻ + 2H₂O + Ni
(c)(i) (6.4 / 24000) ÷ 60 = 4.44 × 10⁻⁶
(mol dm⁻³ s–¹) min 2sf
(ii) [H₂PO₂⁻] doubles / ×2 from experiments 1 to 2 ORA
volume of H₂ produced doubles (∴ first order wrt [H₂PO₂⁻])
[H₂PO₂⁻] ×3 and [OH⁻] ×½ from experiments 1 to 3 ORA
volume of H₂ produced falls to ¾ original
(if first order wrt [H₂PO₂⁻], then must be second order wrt [OH⁻])
ALLOW input data into rate equation and show k is the same
k = 2.8 × 10⁻⁶ / k = 6.7 × 10⁻² (1 / 15) / k = 4 for all experiments [2]
(iii) mol⁻² dm⁶ s⁻¹
(iv) t½ = 0.693 / 8.25 × 10⁻⁵
= 8400 (s) OR t½ = ln2 / 8.25 × 10⁻⁵ = 8401.8 (s)
(v) \((k_1)\) increases (with temperature)
(d) • reactants adsorb (to surface of catalyst)
• bonds (in reactant) weaken
• (reaction occurs and the) products are desorbed
OR reaction occurs and substances are desorbed
any two [1] all three [2]
Questions 3
(a)Topic -28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic – 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(c)Topic -28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(d)Topic -28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
Vanadium is a transition element in Period 4 of the Periodic Table.
(a) Define transition element.
(b) Vanadium shows typical chemical properties of transition elements, including variable oxidation states.
(i) State two other typical chemical properties of transition elements.
(ii) Explain why transition elements have variable oxidation states.
(c) VO₂⁺ can be reduced to V²⁺ by C₂O₄²⁻ in acidic conditions.
equation 2 \(2VO_2^+ + 3C_2O_4^{2–} + 8H^+ \to 2V^{2+} + 6CO_2 + 4H_2O\)
(i) In a titration, 25.00 cm³ of 0.0300 mol/dm³ VO²⁺(aq) is added to 10 cm³ of dilute sulfuric acid. A solution of 0.0400 moldm⁻³ C₂O₄²⁻(aq) is then added from a burette until the end-point is reached. The titration is repeated and concordant results obtained, as shown in Table 3.1.
Show that these results are consistent with the stoichiometry of equation 2.
(ii) An excess of \(C_2O_4^{2–}\) reacts with \(VO_2^+\) to form a mixture of two octahedral complex ions. The complex ions are stereoisomers of each other. Each complex ion contains a \(V^{2+}\) cation and three \(C_2O_4^{2–}\) ions. Complete the diagram to show the three-dimensional structure of one of the complex ions. Include the charge of the complex ion. Use 
to represent a \(C_2O_4^{2–}\) ion
(d) \(V^{2+}\)(aq) can be oxidised by \(H_2O_2\)(aq). Table 3.2 gives some relevant data. 
(i) Identify the vanadium species that forms when an excess of H₂O₂(aq) reacts with V²⁺(aq) under standard conditions. Explain your answer with reference to the data in Table 3.2.
(ii) Concentrated acidified H₂O₂ can react with V²⁺ to form red VO₂³⁺ ions. VO₂³⁺ contains vanadium combined with the peroxide anion, \(O_2^{2–}\). Deduce the oxidation state of vanadium in VO₂³⁺.
▶️Answer/Explanation
Ans:
(a) (a d-block element forms one or more)
stable ions with incomplete filled d-subshell
(b)(i) • they behave as catalysts
• they form complex ions / complexes
• they form coloured compounds / salts / ions
any two
(ii) the d and s sub-shells/orbitals are close/similar in energy
3(c)(i) 0.02500 × 0.0300 OR 7.50 × 10⁻⁴ mol VO₂⁺
OR ½(28.15 + 28.10)/1000 × 0.0400 OR 1.13 × 10⁻³ mol C₂O₄²⁻
Use of their values to show ratio of VO₂⁺: C₂O₄²⁻ = 1:1.5
ALLOW any viable approach
(d)(i) VO₂⁺ AND \(E^0\) of H₂O₂ is largest / most positive value
OR VO₂⁺ AND \(E^0\) is less positive than H₂O₂.
(ii) +5
Questions 4
(a)Topic -34.2 Phenylamine and azo compounds
(b)Topic -34.2 Phenylamine and azo compounds
(c)Topic -36.1 Organic synthesis
(d)Topic -36.1 Organic synthesis
(e)Topic -36.1 Organic synthesis
(f)Topic -36.1 Organic synthesis
Ethylamine and phenylamine are primary amines.
These two compounds are synthesised by different methods.
(a) Several methods can be used to form ethylamine.
(i) Ethylamine forms when ethanamide, CH₃CONH₂, is reduced by \(LiAlH_4\). Write an equation for this reaction. Use [H] to represent one atom of hydrogen from the reducing agent.
(ii) Ethylamine is a product of the reaction of bromoethane with ammonia. Name the mechanism of this reaction and state the conditions used.
mechanism …………………………………………………………………………………………………………
conditions …………………………………………………………………………………………………………..
(iii) The reaction in (a)(ii) also forms secondary and tertiary amines. Suggest the identity of a secondary or tertiary amine formed by the reaction in (a)(ii).
(b) Ethylamine is a weak base. State the relative basicities of ammonia, ethylamine and phenylamine. Explain your answer.
(c) Pure phenylamine, C₆H₅NH₂, can be prepared from benzene in two steps. Draw the structure of the intermediate compound. Suggest reagents and conditions for each step.
(d) Fig. 4.2 shows some reactions of phenylamine.
(i) Draw the structure of W, the organic product of reaction 1.
(ii) State the reagents used in reaction 2.
Benzenediazonium chloride, C₆H₅N₂Cl, and X react together in reaction 4 to form Y, an azo compound.
(iii) Name X, the organic product of reaction 3.
(iv) State the necessary conditions for reaction 4 to occur.
(v) Suggest a use for Y.
(e) Methylamine, CH₃NH₂, is another primary amine. CH₃NH₂ can act as a monodentate ligand.
(i) Define monodentate ligand.
(ii) Cu²⁺(aq) reacts with CH₃NH₂ to form \([Cu(CH_3NH_2)_2(H_2O)_4]^{2+}\). Draw three-dimensional diagrams to show the two geometrical isomers of \([Cu(CH_3NH_2)_2(H_2O)_4]^{2+}\).
(iii) State the coordination number of copper in \([Cu(CH_3NH_2)_2(H_2O)_4]^{2+}\).
(f) Cd²⁺(aq) ions form tetrahedral complexes with \(CH_3NH_2\), \(OH^–\) and \(Cl^–\) ions, as shown in equilibria 1, 2 and 3.
(i) Give the units of \(K_{stab}\) for equilibrium 1.
(ii) Write an expression for \(K_{stab}\) for equilibrium 3.
(iii) A solution of Cl⁻(aq) is added to Cd²⁺(aq) and allowed to reach equilibrium. The equilibrium concentrations are given.
[Cd²⁺(aq)] = 0.043 moldm⁻³
[Cl⁻(aq)] = 0.072 moldm⁻³
Use your expression in (f)(ii) to calculate the concentration of \(CdCl_4^{2–}\)(aq) in the equilibrium mixture.
(iv) When CH₃NH₂(aq) is added to Cd²⁺(aq), a mixture of [Cd(CH₃NH₂)₄]²⁺(aq) and [Cd(OH)₄]²⁻(aq) forms. Suggest how the [Cd(OH)₄]²⁻(aq) is formed.
(v) Cd²⁺(aq) exists as a complex ion, [Cd(H₂O)₆]²⁺(aq). Identify the most stable and the least stable of the complexes in Table 4.1 by placing one tick (3) in each column. Explain your answer.
▶️Answer/Explanation
Ans:
(a)(i) CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O
(ii) nucleophilic substitution
(ammonia with) ethanol AND heat under pressure
OR ethanol AND heat in a sealed tube
(iii) (CH₃CH₂)₂NH OR (CH₃CH₂)₃N
(b) (least) phenylamine < ammonia < ethylamine (most)
explanation
• (order of basicity) ability of base AND to accept a proton
OR donate its lone pair (to a proton)
phenylamine
• lone pair / p-orbital from N delocalised / overlaps with (π-)ring / benzene
ethylamine
• alkyl / ethyl group is electron donating group / +I group
• increases electron density on N (ethylamine) ORA
any two [1] any three [2] all four [3]
(ii) nitrous acid / HNO₂
OR NaNO₂ AND dilute HCl
(iii) phenol
(iv) NaOH / alkali
(v) dyestuffs / dyes
(e)(i) species that uses one lone pair of electrons
that forms a single dative covalent bond to a central metal atom / ion
(iii) 6
(f)(i) units = \(mol^{–4} dm^{12}\)
(ii) \((K_{stab})=\frac{[Cdcl_4^{2-}]}{[Cd^{2+]}{[Cl^-]^4}}\)
(iii) \([CdCl_4^{2–}] = K_{stab} \times 0.043 \times 0.072^4 = 7.28 / 7.3 \times 10^{-4} (mol dm^{-3}) min 2sf\)
(iv) CH₃NH₂ (is basic so) reacts with water to produce \(OH^–\) that reacts with \(Cd^{2+}\)
OR
CH₃NH₂ acts as a base and accepts a proton from Cd[H₂O]₆²⁺ / water
OR
[Cd(H₂O)₆]²⁺ + 4CH₃NH₂ → [Cd(OH)₄(H₂O)₂]²⁻ + 4CH₃NH₃ + OWTTE
(v) Cd(OH)₄²⁻ = most stable (2nd box) AND \([Cd(H_2O)_6]^{2+}\) = least stable (first box)
Cd(OH)₄²⁻ has highest Kstab (and all Kstab values given > 1)
Questions 5
(a)Topic -36.1 Organic synthesis
(b)Topic -36.1 Organic synthesis
(c)Topic -29.4 Isomerism: optical
(d)Topic -29.4 Isomerism: optical
(e)Topic -29.4 Isomerism: optical
(f)Topic -29.4 Isomerism: optical
Tulobuterol is used in some medicines.
(a) Tulobuterol contains a benzene ring in its structure. Describe and explain the shape of benzene.
In your answer, include:
• the bond angle between carbon atoms
• the hybridisation of the carbon atoms
• how orbital overlap forms v and r bonds between the carbon atoms.
(b) In a synthesis of tulobuterol, the first step involves the formation of chlorobenzene. Benzene reacts with Cl₂ in the presence of an AlCl₃ catalyst.
(i) Write an equation to show how Cl₂ reacts with AlCl₃ to generate an electrophile.
(ii) Complete the mechanism in Fig. 5.3 for the reaction of benzene with the electrophile generated in (b)(i). Include all relevant curly arrows and charges. Draw the structure of the intermediate.
(c) The second step of the synthesis involves the reaction of chlorobenzene with \(Cl COCH_2Cl\), also in the presence of an \(Al Cl_3\) catalyst, forming compound Q.
(i) Name the mechanism of the reaction in step 2.
(ii) Draw the structure of an isomer of Q that forms as an organic by-product of the reaction in step 2.
(iii) The reactants used in step 2 contain acyl chloride, alkyl chloride and aryl chloride functional groups. State and explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and aryl chlorides.
(d) Tulobuterol is produced from Q as shown in Fig. 5.5.
Suggest reagents and conditions for steps 3 and 4. Draw the structure of Z in the box.
step 3 ……………………………………………………………………………………………………………………….
step 4 ……………………………………………………………………………………………………………………….
(e) The synthesis produces two enantiomers of tulobuterol.
(i) Define enantiomers.
(ii) Suggest one disadvantage of producing two enantiomers in this synthesis.
(iii) Suggest a method of adapting the synthesis to produce a single enantiomer.
(f) (i) Predict the number of peaks that would be seen in the carbon-13 NMR spectrum of tulobuterol.
(ii) The proton \((^1H)\) NMR spectrum of tulobuterol dissolved in \(D_2O\) shows peaks in four different types of proton environment. The peak for the —CH₂N— environment is a doublet in the chemical shift range \( \delta = 2.0–3.0ppm\). Give details for each of the other three peaks in the proton NMR spectrum of tulobuterol, to include:
• chemical shift
• environment of the proton
• splitting pattern
• number of \(^1H\) atoms responsible.
Table 5.1 gives information about typical chemical shift values.
▶️Answer/Explanation
Ans:
a) any three points from:
• bond angle = 120°
AND shape is hexagonal ring planar / trigonal planar
• (carbons are) \(sp^2\) hybridised
• contains delocalised electrons in the π bonds / system
• (sp2 orbitals) overlap end-on-end/ head-on to form σ bonds
• a p orbital (from each carbon atom) overlaps sideways
(with each other above and below the ring) forming π bonds
(b)(i) Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻
(c)(i) electrophilic substitution
(iii) (most) acyl chloride > alkyl chloride > aryl chloride (least)
any two from:
• acyl chlorides: carbon (in C—Cl) is more electron deficient
AND it is also attached to an oxygen atom / two electronegative atoms
OR C—Cl bond is weakest / weakened in acyl chlorides
AND it is also attached to an oxygen atom / two electronegative atoms
• aryl chlorides (no hydrolysis) C—Cl bond is part of delocalised system / partially double bond character so C—Cl
bond is stronger
OR lone pair on Cl delocalises with the π ring so C—Cl bond is stronger
• alkyl chlorides carbon atom has a smaller δ+
AND due to the carbon being only attached to one electronegative atom
OR C—Cl bond strengthened
AND by electron donating effect / positive inductive effect of alkyl / R group
(e)(i) rotate the plane of polarised light in the opposite direction
OR stereoisomers / molecules that are non-superimposable mirror images
(ii) need to separate the optical isomers to form the pure active isomer
OR reduced / different biological activity of ‘other’ enantiomer
OR lower yield of biologically active molecule / desired molecule
(iii) chiral catalyst OR use of an enzyme
(f)(i) ten / 10
(ii) (δ = 0.9–1.7)
• 9H
• singlet
• –CH₃ / alkane
(δ = 3.2–4.0) / (δ = 2.3–3.0)
• 1H
• triplet
• –CHO / alkyl next to electronegative atom
OR Ar-CH / alkyl next to aromatic ring
(δ = 6.0–9.0)
• 4H
• multiplet
• H–Ar / attached to aromatic ring
any three [1] any six [2] all nine [3]
Questions 6
(a)Topic -37.2 Gas/liquid chromatography
(b)Topic -37.2 Gas/liquid chromatography
(c)Topic -37.2 Gas/liquid chromatography
(d)Topic -37.2 Gas/liquid chromatography
A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents. Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent.
(a) (i) Suggest a compound that could be used as the stationary phase in this experiment.
(ii) Table 6.1 shows the \(R_f\) values for different metal cations when separated by TLC using water as a solvent. 
Suggest the identity of the cation that causes the spot at M in Fig. 6.1. Explain your answer.
(b) The student repeats the experiment using butan-1-ol as a solvent. The metal cations do not travel as far up the TLC plate in this experiment. Suggest why the metal cations do not move as far up the TLC plate with butan-1-ol as a solvent.
(c) The student sprays the TLC plate in Fig. 6.1 with KSCN(aq). The colour of some of the spots changes, as some of the metal cations undergo a ligand exchange reaction. Identify the ligands involved in the ligand exchange reaction.
(d) In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution. The student prepares the buffer solution by mixing 20.0 cm³ of 0.150 mol/dm³ KOH(aq) and 50.0 cm³ of 0.100 mol/dm³ C₈H₅O₄K(aq). \(C_8H_5O_4K\) is a weak carboxylic acid that has \(pK_a = 5.40\).
(i) Complete the equation for the reaction of C₈H₅O₄K(aq) with KOH(aq).
(ii) Calculate the pH of the buffer solution. Show all your working.
▶️Answer/Explanation
Ans:
(a)(i) SiO₂ OR Al₂O₃ OR silica OR alumina OWTTE
(ii) Cd²⁺ AND \(R_f\) of M (= 2/5) = 0.40 / same \(R_f\) as in the Table 6.1
(b) metal cations are less soluble in butan-1-ol (than in water)
OR metal cations weaker ion-dipole forces with butan-1-ol
(c) H₂O AND \(SCN^–\)
